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3 years ago

Please help! 10 points

Mathematics

2 answers:

Roman55 [17]3 years ago

5 0

Answer:

-80 answer = -80 80 80 80 80

Svetach [21]3 years ago

4 0

-80 ?

................

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What is the arc length of a circle that has a 6-inch radius and a central angle that is 65 degrees? Use 3. 14 for pi and round y

noname [10]

C=2πr

C=2π6

C=37.6991118431

That's the arc length of the whole circle, i.e. the arc length of 360°.

65° is 18.055555555555555555555555555556% of 360°, (65/360*100)
so 18.055555555555555555555555555556% of 37.6991118431 is
6.8067840827819444444444444444444. Rounded to the nearest hundredth, the answer is 6.81 inches.

That's real pi, lets see if it makes a difference to use "stupid pi".

C=2πr

C=2π6

C=37.68

That's the arc length of the whole circle, i.e. the arc length of 360°.

65° is 18.055555555555555555555555555556% of 360°, (65/360*100)
so 18.055555555555555555555555555556% of 37.68 is
6.803333333333333333333333. Rounded to the nearest hundredth, the answer is 6.80 inches.
Yep makes a difference. That's why you don't use stupid pi. 3.14159 is what we always used in engineering, or just the pi button and using a ton of digits.

Answer: 6.80 inches.

3 0

3 years ago

Please help I'm not the best at word problems and I really need help

k0ka [10]

Answer :
70mins(T) x 65(d) divided by half

You have 85miles left so 1hour and 30mins

5 0

3 years ago

The perimeter of a rectangular field is 344m.if the length of the field is 93m.what is it’s width?

snow_tiger [21]

Answer:

79m

Step-by-step explanation:

344-(93x2)=158

158/2=79m

4 0

2 years ago

Read 2 more answers

Any 10th grader solve it for 50 points

kkurt [141]

Answer:

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)\neq 0$ is proved for the sum of pth, qth and rth terms of an arithmetic progression are a, b,and c respectively.

Step-by-step explanation:

Given that the sum of pth, qth and rth terms of an arithmetic progression are a, b and c respectively.

First term of given arithmetic progression is A

and common difference is D

ie., $a_{1}=A$ and common difference=D

The nth term can be written as

$a_{n}=A+(n-1)D$

pth term of given arithmetic progression is a

$a_{p}=A+(p-1)D=a$

qth term of given arithmetic progression is b

$a_{q}=A+(q-1)D=b$ and

rth term of given arithmetic progression is c

$a_{r}=A+(r-1)D=c$

We have to prove that

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)=0$

Now to prove LHS=RHS

Now take LHS

$\frac{a}{p}\times (q-r)+\frac{b}{q}\times (r-p)+\frac{c}{r}\times (p-q)$

$=\frac{A+(p-1)D}{p}\times (q-r)+\frac{A+(q-1)D}{q}\times (r-p)+\frac{A+(r-1)D}{r}\times (p-q)$

$=\frac{A+pD-D}{p}\times (q-r)+\frac{A+qD-D}{q}\times (r-p)+\frac{A+rD-D}{r}\times (p-q)$

$=\frac{Aq+pqD-Dq-Ar-prD+rD}{p}+\frac{Ar+rqD-Dr-Ap-pqD+pD}{q}+\frac{Ap+prD-Dp-Aq-qrD+qD}{r}$

$=\frac{[Aq+pqD-Dq-Ar-prD+rD]\times qr+[Ar+rqD-Dr-Ap-pqD+pD]\times pr+[Ap+prD-Dp-Aq-qrD+qD]\times pq}{pqr}$

$=\frac{Arq^{2}+pq^{2} rD-Dq^{2} r-Aqr^{2}-pqr^{2} D+qr^{2} D+Apr^{2}+pr^{2} qD-pDr^{2} -Ap^{2}r-p^{2} rqD+p^{2} rD+Ap^{2} q+p^{2} qrD-Dp^{2} q-Aq^{2} p-q^{2} prD+q^{2}pD}{pqr}$